Consider a subset $U \subset \mathbb{R}^{n}$. Clearly, $U$ can be considered as a smooth ($n$-dimensional) submanifold of $\mathbb{R}^{n}$. A Riemannian metric on $U$ is a smooth map $g$ which associates to each point $p \in U$ an inner product $\left\langle \cdot,\cdot \right\rangle_{p}$ on $T_{p}M$.
In this context, for each $p \in U$, the tangent space $T_{p}M$ can be identified with $\mathbb{R}^{n}$. An inner product on $\mathbb{R}^{n}$ is, necessarily of the form $(u,v) \to u^{\top}Av$ where $A$ is a positive definite matrix.
My question (a bit naive) is the following : are all the Riemannian metrics on $U$ of the form $g : p \in U \mapsto g_{p}$ with : $g_{p}(u,v) = u^{\top}A(p)v$ for all $(u,v) \in T_{p}M$ ?
After a quick search, I think the answer is no. I believe that the Poincaré half-plane provides a counter example. However, is it true if $n=1$ ? Are there results regarding the "structure" of Riemannian metrics ?
Firstly, $U$ has to be open to be an $n$-dimensional submanifold of $R^n$. If $g$ is a differentiable metric, then for every $p\in U$, $g_p$ is a scalar product defined on $T_pU=R^n$. So the answer is yes.